Hi,
I think the solution given is not correct (A^B)^A does not give A.B
we get the following o/p
A B o/p
0 0 0
0 1 1
1 0 0
1 1 1
Can anyone provide the correct answer ?
My apologies, wasnt thinking too well. Anyway, this should work, although I'm not too sure if it uses the least XOR gates. Baiscally uses the idea of !A = 1^A.
A.B = (((A^B)^1)^A)^1
Any combinational circuit can be accomplished using NAND/NOT or NOR/NOT gates. THe idea is to manipulate using De Morgan's Theorem.
By boolean algebra, (0,1,&,|) forms a field, (0,1,^,|) also forms a field. use de morgan dicipline, add ~ operator, (&, ~) is enough, (| ~) is also enough. but ^ can't benefit from de morgan, (^, ~) is not enough.
In this example, I think A.B can't be represented by only ^ operator.
Yes!
You can't have AND gate using XOR gate! You can only get
buffer or inverter!
To be an universal gate a gate needs to satisfy both the following conditions!
1. You should be able to get inverter using the gate.
2. You should be able to block the input.
Here by 2 what i means is this....
Take a 2 input NAND gate
1. is satisfied if you connect one input to '1' o/p will be inversion of input!
2. is satisfied if you connect one input to '0' o/p will be '1' other input wont
have any effect on o/p. i.e. other input gets blocked!
Only a MUX, NAND and NOR gates satisfy both the above conditions and are
Universal gates!
XOR satisfies only 1st condition!
Any combinational logic can be implemented using a universal gate. A nand gate and a nor gate are called as universal gate. (U even dont require a not gate). U can also implement a combinationla logic or basic gates using muxes. Implementing comb logic using muxes is useful in the cases of FPGAs, if you are having a shortage of comb gates.